Module Content:

Module Reading, Problems, and Demo:

MAE 2310 Str. of Materials © E. J. Berger, 2010 15- 1

Module 15: Composite BeamsMarch 17, 2010

1. Composite beams are composed of two or more different materials bonded together to form a beam. The bending analysis can be treated using a transformation factor which is the ratio of the moduli of the two materials.2. Reinforced concrete beams use a similar transformation in bending analysis, but the transformation is less physical because the concrete is assumed to support no load in tension.

Reading: Sections 6.6, 6.7Problems: Example 6.21, Example 6.23, Prob. 6-123Demo: noneTechnology: http://pages.shanti.virginia.edu/som2010

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Composite Beams• a composite beam consists of two materials perfectly bonded along an interface

• as we shall see, the neutral axis of bending (zero strain) does NOT correspond to the geometric center of the cross section (the centroid)

• the strain field is continuous, as is required by continuity conditions on the material behavior

• however, because of the modulus mismatch, the stress field is discontinuous

2

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Composite Beams• a composite beam consists of two materials perfectly bonded along an interface

• as we shall see, the neutral axis of bending (zero strain) does NOT correspond to the geometric center of the cross section (the centroid)

• the strain field is continuous, as is required by continuity conditions on the material behavior

• however, because of the modulus mismatch, the stress field is discontinuous

2

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Composite Beams• a composite beam consists of two materials perfectly bonded along an interface

• as we shall see, the neutral axis of bending (zero strain) does NOT correspond to the geometric center of the cross section (the centroid)

• the strain field is continuous, as is required by continuity conditions on the material behavior

• however, because of the modulus mismatch, the stress field is discontinuous

2

N.A.

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Composite Beams• a composite beam consists of two materials perfectly bonded along an interface

• as we shall see, the neutral axis of bending (zero strain) does NOT correspond to the geometric center of the cross section (the centroid)

• the strain field is continuous, as is required by continuity conditions on the material behavior

• however, because of the modulus mismatch, the stress field is discontinuous

2

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Composite Beams• a composite beam consists of two materials perfectly bonded along an interface

• as we shall see, the neutral axis of bending (zero strain) does NOT correspond to the geometric center of the cross section (the centroid)

• the strain field is continuous, as is required by continuity conditions on the material behavior

• however, because of the modulus mismatch, the stress field is discontinuous

2

!1 = E1"

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Composite Beams• a composite beam consists of two materials perfectly bonded along an interface

• as we shall see, the neutral axis of bending (zero strain) does NOT correspond to the geometric center of the cross section (the centroid)

• the strain field is continuous, as is required by continuity conditions on the material behavior

• however, because of the modulus mismatch, the stress field is discontinuous

2

!2 = E2"

!1 = E1"

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Transformation Factor• we can develop a mapping from the two-material problem to a one-material problem (so that we can use the

flexure formula directly) through a simple geometric scaling

• we alter the width of the one of the materials so that the load carrying capacity of the transformed material remains the same using the transformation factor n, defined as:

3

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Transformation Factor• we can develop a mapping from the two-material problem to a one-material problem (so that we can use the

flexure formula directly) through a simple geometric scaling

• we alter the width of the one of the materials so that the load carrying capacity of the transformed material remains the same using the transformation factor n, defined as:

3

n =

E1

E2

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Transformation Factor• we can develop a mapping from the two-material problem to a one-material problem (so that we can use the

flexure formula directly) through a simple geometric scaling

• we alter the width of the one of the materials so that the load carrying capacity of the transformed material remains the same using the transformation factor n, defined as:

3

n =

E1

E2

n!=

E2

E1

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Stress Calculations• once we implement the transformation factor, we can calculate the stresses on the transformed geometry

exactly as before using the flexure formula:

• then on the transformed section:

4

1. find the centroid2. find the moment of inertia3. determine the bending stress

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Stress Calculations• once we implement the transformation factor, we can calculate the stresses on the transformed geometry

exactly as before using the flexure formula:

• then on the transformed section:

4

1. find the centroid2. find the moment of inertia3. determine the bending stress

n =

E1

E2

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Stress Calculations• once we implement the transformation factor, we can calculate the stresses on the transformed geometry

exactly as before using the flexure formula:

• then on the transformed section:

4

1. find the centroid2. find the moment of inertia3. determine the bending stress

n =

E1

E2

n!=

E2

E1

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

original problem: composite beam

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

transformed section:

btr =12 GPa

200 GPa(150) = 9 mm

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

stress profile on transformed section

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

stress profile on original section

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Final Stress Transformation• by example: Example 6.21

5

stress profile on original section

!w = n!tr

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Concept: Reinforced Concrete Beams• concrete beams are easy to manufacture and cost effective to use, but their strength in tension is unsatisfactory

(for reasons that you should already know)

• as such, concrete beams are typically reinforced by steel bars on the tension side of the neutral axis, resulting in a composite beam with “point” inhomogeneities (i.e., the second material is surrounded by the first)

• how do we analyze this incredibly important and relevant type of composite beam?

6

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Steel Reinforcements• we go back to this notion that the load carrying capacity of any transformed section should be the same as the

original section, but here we do two things

• replace the steel reinforcements with a larger area of concrete (n > 1)

• position this new concrete section precisely where the steel reinforcements are located

7

n =

Est

Econc

!

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Remarks and Mathematics• the transformed steel section has no physical

dimensions, it is a transformed area nAst

• the transformed section is located at a distance d from the top of the beam, and this is the same as the distance from the top to the centerline of the steel reinforcements

• the centroid h’ of the new section is calculated the usual way:

8

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Remarks and Mathematics• the transformed steel section has no physical

dimensions, it is a transformed area nAst

• the transformed section is located at a distance d from the top of the beam, and this is the same as the distance from the top to the centerline of the steel reinforcements

• the centroid h’ of the new section is calculated the usual way:

8

bh!

!

h!

2

"

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Remarks and Mathematics• the transformed steel section has no physical

dimensions, it is a transformed area nAst

• the transformed section is located at a distance d from the top of the beam, and this is the same as the distance from the top to the centerline of the steel reinforcements

• the centroid h’ of the new section is calculated the usual way:

8

bh!

!

h!

2

"

!nAst(d ! h!)

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Remarks and Mathematics• the transformed steel section has no physical

dimensions, it is a transformed area nAst

• the transformed section is located at a distance d from the top of the beam, and this is the same as the distance from the top to the centerline of the steel reinforcements

• the centroid h’ of the new section is calculated the usual way:

8

bh!

!

h!

2

"

!nAst(d ! h!)= 0

MAE 2310 Str. of Materials © E. J. Berger, 2010 15-

Theory: Remarks and Mathematics• the transformed steel section has no physical

dimensions, it is a transformed area nAst

• the transformed section is located at a distance d from the top of the beam, and this is the same as the distance from the top to the centerline of the steel reinforcements

• the centroid h’ of the new section is calculated the usual way:

8

bh!

!

h!

2

"

!nAst(d ! h!)= 0

b

2h!2

+ nAsth!! nAstd = 0